Prove that if you draw a right triangle and then draw a circle with its center on the hypotenuse's midpoint such that it intersects at least 1 vertex, it will in fact intersect all three.

Consider the right triangle ABC with angle C as the right angle. Let D be the midpoint of the hypotenuse. Draw the circle with center D and radius equal to one-half of the hypotenuse. By definition, this circle contains points A and B. So the proof is to show that C is on the circle.

First of all, measure of any inscribed angle of a circle is half the measure of the arc. If C were on the circle, it would have a measure of one-half of the arc AB. Since AB is a diameter, arc AB has a measure of 180, so angle C would have to be a right angle if it were on the circle.

To prove the converse, we use an indirect proof. Assuming that C is not on the circle. We could find a point P on the line formed by AC (or BC) that intersects the circle. Then angle APB must be a right angle by the above. However, you can only draw ONE unique perpendicular through one point (point B in this case) to a given line (AC in this case). Hence, this is a contradiction, which means that C must be on the circle.

Since A, B, and C are on the circle, they are all equidistant from the midpoint of the hypotenuse. Therefore, drawing any circle with the center on the midpoint of the hypotenuse and containing at least one of the vertices will contain all three.

QED

Posted by np_rt on 2003-11-19 13:08:47